Quantitative
Composition of
Compounds
•Define the MOLE
•Determine molar mass of compounds
•Calculate percent composition of compounds
•Distinguish the differences between (and be able
to calculate) empirical formulas and molecular
formulas
Composition of Compounds
• Ratios of ingredients determines end result
– Making cookies – can I double the salt?
• Ratios in chemical compounds
– Determine identity of the substance
– Have specific properties…
– Changing changes the entire compound
– Example
• H2O versus H2O2 –does one extra oxygen really
make a difference???
Chemical Measurements
• How can we measure masses of atoms?
– Atomic mass units too small to use on
scale
– If we use bigger amounts, how do we
know how many atoms are in it?
• Counting by weighing
– Know average mass of one
– Weigh sample, divide by average mass of
one
Chemical Measurements
• Atomic mass:
– The averages of isotopes for a naturally
occurring element, expressed in atomic
mass units (amu).
• Mole (mol):
– The quantity of a given substance that
contains as many molecules (formula units)
as the number of atoms in exactly 12 g of
carbon-12
Chemical Measurements
• Avogadro’s number:
– Number of items in a mole
– Value is 6.022 x 10-23
• Molar mass:
– The mass (in grams) of one mole of any
substance
– Molar mass (g) = formula or atomic mass
(amu)
The Mole
• Mole – unit for counting atoms
– Measures a mass, a number of particles, and a
volume of a gas
– Amadeo Avagadro (1776-1856)
• Studied quantitative chemistry
– Avagadro’s number
• 6.022 x 1023
• Number of particles (atoms, molecules, anything) in one
mole
• Constant for ANY SUBSTANCE
The Mole
• Molar Mass
– Grams of element (or compound) that
contain Avogadro’s number of atoms
– H = 1.008 g
– Mg = 24.31 g
– O = 15.9994 g
– Simply take atomic mass (in amu) and
“convert” to grams
The Mole
• The atomic mass expressed in grams is
the molar mass of an element
– Express to four significant figures
– Atomic mass of Flourine is 18.99 amu
– Molar mass of Flourine is 18.99 g
• One mole of any element contains
Avagadro’s number of atoms
– 1 mol = 6.022 X1023 atoms
The Mole
• Can now convert between:
mass
moles
molecules
• 25.0 g of iron is how many moles?
25.0 g 1 mol
= 0.448 mol
55.85 g
The Mole
• How many grams of Na are in 2.55 moles?
2.55 mol 22.99 g =
58.6 g
1 mol
• How many atoms of Carbon are in 2.35
grams?
2.35 g 1 mol 6.022 X 1023 atoms =1.18 x 10-23
atoms
12.0 g
1 mol
Molar Mass of Compounds
• One mole of a compound contains
6.022 x 10-23 formula units of that
compound
– Formula unit – molecule, atom, etc.
– One mole of H2O contains 6.022 x 10-23
molecules of H2O
Molar Mass of Compounds
• Molar mass – add molar masses of all
atoms
– What is the molar mass of H2O?
• 2 hydrogen + 1 oxygen
• 2 (1.008 g) + 1 (16.00 g) = 18.02 g
– What is the molar mass of Ca(OH)2?
• 1(40.08g) +2(16.00g) + 2(1.008g) = 74.10 g
Molar Mass of Compounds
• How many molecules are there in a
3.46 g sample of HCl?
3.46 g HCl
1 mol HCl
36.5 g HCl
6.02 x 1023 molecules HCl
1 mol HCl
= 5.71 x 1022 molecules
Molar Mass of Compounds
• If you burned 6.10 x 1024 molecules of
ethane (C2H6) what mass of ethane did
you burn?
6.10 x 1024 molecules
1 mol C2H6
30.07 g C2H6
6.02 x 1023 molecules 1 mol C2H6
= 304.7 g C2H6
Moles and Gases
• At the same temperature and pressure,
equal volumes of gases contain the
same number of gas particles
• STP = Standard Temperature (00 C) and
Pressure (1 atmosphere)
• 1 mole of any gas at STP = 22.4 L
– Molar volume
Moles and Gases
• A student fills a 1.0 L flask with CO2 at
STP. How many molecules of gas are in
the flask?
1.0 L CO2
1 mol CO2 6.02 x 1023 molecules
22.4 L CO2
1 mol CO2
= 2.7 x 1022 molecules CO2
Moles and Gases
• A container with a volume of 893 L contains how
many moles of air at STP?
893 L air
1 mol air
= 39.9 mol air
22.4 L air
• A chemical reaction produces 0.37 moles of N2
gas. What volume will that gas occupy at STP?
0.37 mol N2
22.4 L N2
1 mol N2
= 8.3 L N2
Percentage Composition of
Compounds
• Mass of one element in a compound
compared to the total mass of the
whole compound
• Two ways to determine…
– Calculate using given chemical formula
– Experimental analysis
• Use data from experiment to calculate
Percentage Composition of
Compounds – from Formula
1) Calculate the molar mass
2) Divide the total mass of each element
in the formula by the molar mass and
multiply by 100.
Total mass of element x 100 = % of element
molar mass
Percentage Composition of
Compounds – from Formula
• What is the percent composition of
each element in table salt (NaCl)?
1 mol NaCl = 1 mol Na + 1 mol Cl
58.44 g
= 22.99 g
35.45 g
23 g Na
x 100% = 39.65% Na
58 g NaCl
35 g Cl
x 100% = 60.34% Cl
58 g NaCl
Should
equal
100%
Percentage Composition of
Compounds – from
Experimental Data
1) Calculate the mass of the compound
formed
2) Divide the mass of each element by
the total mass of the compound and
multiply by 100
Percentage Composition of
Compounds
• A sample of 2.45 g of aluminum oxide
decomposes into 1.30 g of aluminum and 1.15
g of oxygen. What is the percent composition
of each element?
Percentage Al = 1.30 g x 100%
2.45 g
=53.1%
Percentage O = 1.15 g x 100%
2.45 g
=46.9%
Percentage Composition of
Compounds
• Find the percentage composition of a
compound that contains 1.94 g of
carbon, 0.480 g of hydrogen, and 2.58
g of sulfur in a 5.00 g sample of that
compound.
38.8 %C
9.60% H
51.6% S
Empirical Formula
• A formula that gives the SIMPLEST
whole-number ratio of elements in that
compound
Ratio of masses
(percent composition)
Ratio of atoms
(formula)
Empirical Formula
• What is the empirical formula of a compound containing
40% carbon, 53.3% oxygen, and 6.7% hydrogen?
In a 100 g sample there would be 40 g C, 53.3 g O, and 6.7 g H
40 g C 1 mol C = 3.33 mol C = 1
12.0 g C
3.33
53.3 g O 1 mol O = 3.33 mol O = 1
16.0 g O
3.33
6.7 g H 1 mol H = 6.7 mol H = 2
3.33
1.0 g H
Divide by
smallest
number
Formula is:
COH2
Empirical Formula
• Determine the empirical formula of a
compound containing 5.75 g Na, 3.5 g
N, and 12.0 g O
– Find percent composition FIRST
– Then, use percents to calculate # of moles
– Then, divide by smallest number
– Write formula
NaNO3
Empirical Formula
• Determine the empirical formula of a
compound containing 2.644g of gold
and 0.476 g of chlorine.
AuCl
Molecular Formula
• Formula that gives the actual number of
atoms in a given compound
• Some compound may have same empirical
formula but actually be very different
– Acetylene vs Benzene
• Sometimes emipirical formula and molecular
formula are the same
– Water
• Compare molar mass with empirical formula
mass
Molecular Formula
• Ribose has a molar mass of 150 g/mol and a
chemical composition of 40.0% carbon,
6.67% hydrogen, and 53.3% oxygen. What
is the molecular formula for ribose?
–
–
–
–
Find empirical formula
Calculate empirical formula mass
Compare to molecular formula mass
Multiply empirical formula by difference
C5H10O5
Molecular Formula
• Find the molecular formula of a compound
that contains 42.56 g of palladium and 0.80 g
of hydrogen. The molar mass of the
compound is 216.8 g/mol.
• Octane, a compound of hydrogen and carbon,
has a molar mass of 114.26 g/mol. If the
compound contains 18.17 g/mol hydrogen,
what is its molecular formula?
Homework
• Questions #1, 6-10
• Paired Exercises #11-31 odd (a, b, c only);
33; 37-41 odd
• Additional Exercises #43, 47, 50 & 51
• NO CLASS TUESDAY – Study help???
– Morning, afternoon, evening?
• TEST THURSDAY – Chapters 11, 6 & 7